Great thanks very much - when I add E3 and E4 I get a quadratic equation in lambda which gives lambda =1 or lambda = -2. So ruling out lambda=1 which gives an infinite number of solutions then lambda = -2. Would I be right? Thanks again

Great thanks very much - when I add E3 and E4 I get a quadratic equation in lambda which gives lambda =1 or lambda = -2. So ruling out lambda=1 which gives an infinite number of solutions then lambda = -2. Would I be right? Thanks again

We need to be vary careful right now.

Both both give an infinite number of solutions.

Back in my first post I recommended that you muliply one of the equations by

right?

What value can't lambda be? Asked another way when we multiply an equation by a number can we use any number or are there number(s) we can use?

If you are lost, it might be best to go back and think about what you are doing. You originally said, "Find lambda for the linear system", and then gave a system of equations involving x, y, z and parameter lambda. Find lambda so that what happens? I suspect you want to find lambda so that the system has a unique solution. Or, the contrary- find lambda so that there is more than one solution. Why would that be of interest? I can't imagine why you would say "The simultaneous equation part didn't seem to help this atall?" Without the system of equations, you would only be saying "Find lambda" without any conditions on it. And that would make no sense at all!